Question

Solve 2x – cosx – 3 = 0 by using the method of successive approximations correct of three decimal places.

Answer 1

1.523

Answer 2

Let the given equation be $f(x) = 2x - \cos x - 3 = 0$.

We use the Newton-Raphson method: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.

$f'(x) = 2 + \sin x$.

The iteration formula is $x_{n+1} = x_n - \frac{2x_n - \cos x_n - 3}{2 + \sin x_n}$.

Evaluate $f(1) \approx -1.54$ and $f(2) \approx 1.42$. A root exists between 1 and 2.

Choose initial guess $x_0 = 1.5$.

Iterate:

$x_1 \approx 1.5 - \frac{f(1.5)}{f'(1.5)} \approx 1.5 - \frac{-0.0707}{2.9975} \approx 1.5236$.

$x_2 \approx 1.5236 - \frac{f(1.5236)}{f'(1.5236)} \approx 1.5236 - \frac{0.000897}{2.9990} \approx 1.5233$.

$x_3 \approx 1.5233 - \frac{f(1.5233)}{f'(1.5233)} \approx 1.5233 - \frac{0.000000}{2.9991} \approx 1.5233$.

The value converges to $1.5233$.

Rounding to three decimal places gives $1.523$.

Verify $f(1.523) \approx -0.00081$, which is close to zero.

The root correct to three decimal places is $1.523$.